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7/31/2019 Differential Equations Review
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7/31/2019 Differential Equations Review
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In the following we will BRIEFLY review the basics of solving Linear, Constant
Coefficient Differential Equations under the Homogeneous Condition
Homogeneous means the forcing function is zero
That means we are finding the zero-input response that occurs due to the effect
of the initial coniditions.
Write D.E. like this:
) ) )(...)(...)(
01
)(
01
1
1 tfbDbDbtyaDaDaD
DP
m
m
DQ
n
n
n
444 3444 2144444 344444 21==
+++=++++
.. EqDiff
)()()()( tfDPtyDQ =
m is the highest-order derivative
on the input side
n is the highest-order derivative
on the output side
We will assume: m n
)()( tyDdt
tyd kk
k
Use operational notation:
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Puty0(t) into () and use result for the derivative of an exponential:
0)...( 011
1=++++
tn
n
n
eaaac
must = 0
solutionais
solutionais
solutionais
2
1
2
1
t
n
t
t
n
ec
ec
ec
M
Then, choose c1, c2,,cn to satisfy the given ICs
t
n
tt
zinecececty
+++= ...)(:SolutionI-Z 21 21
tn
n
tn
edt
ed
=
Characteristic polynomial
Q() has at most n unique roots
(can be complex)
))...()(()(21 n
Q =
Soany linear combination
is also a solution to ()
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{ }nitie 1= Set of characteristic modes
Real Root: tii
iej
+= 0
real real t
0>i
0=i0
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To get only real-valued solutions requires the system coefficients to be
real-valued.
Complex roots of C.E. will appear in conjugate pairs:
j
j
k
i
=
+=Conjugate pair
tjt
k
tjt
i
t
k
t
i eeceecececki +=+
For some real Cj
eC
2j
eC
2
0)cos( >+ ttCet
Use Euler!
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Repeated Roots
Say there are rrepeated roots
))...()(()()( 321 pr
Q = p = n - r
ZI Solution:
:modesother...)( 111211 ++++= tr
rziietctccty
effect ofr-repeated roots
See examples on the next several pages
We can verify that: trttt etettee i 111 12 ,...,,, satisfy ()
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Find the zero-input response (i.e., homogeneous solution) for
these three Differential Equations.
Example (a)
)()(2)(3)(
)()(2
)(3
)(
2
2
2
tDftytDytyD
dt
tdfty
dt
tdy
dt
tyd
=++
=++
The zero-input form is:
0)(2)(3)(
0)(2)(
3)(
2
2
2
=++
=++
tytDytyD
tydt
tdy
dt
tyd
The Characteristic Equation is:
0)2)(1(0232 =++=++
Differential Equation Examples
w/ I.C.s
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The Characteristic Equation is:
0)2)(1(0232 =++=++
2&1 21
The Characteristic Roots are:
==
The Characteristic Modes are:
tttteeee
221 & ==
The zero-input solution is:
ttzi eCeCty
221)(
+=
The System forces this
form through its Char. Eq.
The ICs determine the
specific values of the Cis
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The zero-input solution is:
ttzi eCeCty
221)(
+=
and it must satisfy the ICs so:
0)0(0 210
20
1 =++==
CCeCeCyzi
The derivative of the z-s soln. must also satisfy the ICs so:
522)0(5 21
0
2
0
1=+==
CCeCeCyzi
Two Equations in Two Unknowns leads to:
5&5 21==
CC
The particular zero-input solution is:
321321modesecond
2
modefirst
55)( ttzi eety +=
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Example (b):
)(5)(3)(9)(6)(
)(5)(
3)(9)(
6)(
2
2
2
tftDftytDytyD
tfdt
tdfty
dt
tdy
dt
tyd
+=++
+=++
The zero-input form is:
0)(9)(6)(
0)(9)(
6)(
2
2
2
=++
=++
tytDytyD
tydt
tdy
dt
tyd
The Characteristic Equation is:
0)3(096 22 =+=++
w/ I.C.s
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The Characteristic Equation is:
The Characteristic Roots are:
3&3 21 ==
The Characteristic Modes are:
ttttteteee
33 21 & ==
The zero-input solution is:
ttzi teCeCty
32
31)(
+=
The System forces this
form through its Char. Eq.
The ICs determine the
specific values of the Cis
0)3(096 22 =+=++
Using the rule to
handle repeated roots
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Following the same procedure (do it for yourself!!) you get
The particular zero-input solution is:
tttzi etteety
3
modesecond
3
modefirst
3 )23(23)( +=+= 321321
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0 1 2 3 4 5 6 7 8 9 100
1
2
3
FirstMode
Plots for Example 2.1 (b)
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
SecondMo
de
0 1 2 3 4 5 6 7 8 9 100
1
2
3
t (sec)Zero-InputResponse
Plots for Example (b)
Same Decay Rates
Effect of t out in front
Because the characteristic roots are real and negative
the modes and the Z-I response all .
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Example (c):
)(2)()(40)(4)(
)(2)(
)(40)(
4)(
2
2
2
tftDftytDytyD
tfdt
tdfty
dt
tdy
dt
tyd
+=++
+=++
The zero-input form is:
0)(40)(4)(
0)(40)(
4)(
2
2
2
=++
=++
tytDytyD
tydt
tdy
dt
tyd
The Characteristic Equation is:
0)62)(62(04042 =+++=++ jj
w/ I.C.s
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The Characteristic Equation is:
The Characteristic Roots are:
62&62 21 jj =+=
The Characteristic Modes are:
tjtttjtteeeeee
6262 21 & + ==
The zero-input solution is:
tjttjtzi eeCeeCty
622
621)(
+ +=
The System forces this
form through its Char. Eq.
The ICs determine the
specific values of the Cis
0)62)(62(04042 =+++=++ jj
F ll i h d i h i l i f
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Following the same procedure with some manipulation of
complex exponentials into a cosine
The particular zero-input solution is:
)3/6cos(4)( 2 += tety tzi
Set by the ICs
Imag. part of root
controls oscillation
Real part of root
controls Decay
l x
Plots for Example (c)
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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1
Firs
tMode
Plots for Example 2.1 (c)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1
SecondMode
0 0.5 1 1.5 2 2.5 3 3.5 4-4
-2
0
2
4
t (sec)Zero-InputR
esponse
CantEasilyPlot
theModes
BecauseTheyA
reComple
xPlots for Example (c)
Because the characteristic roots are complex have oscillations!
Because real part of root is negative !!!
4e-2t
-4e-2t
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Big Picture
The structure of the D.E. determines
the char. roots, which determine the
character of the response:
Decaying vs. Exploding (controlled by real part of root) Oscillating or Not (controlled by imag part of root)
The D.E. structure is determined by the
physical systems structure and
component values.